Question

A ball whose kinetic energy is E is thrown at an angle of 45° with horizontal. Its kinetic energy at highest point of flight will be(a) E (b) $\frac{E}{2}$(c) $\frac{E}{\sqrt{2}}$(d) O

Answer 1

at the highest point, Vy will become zero. hence, the velocity attained by the body will only be due to its horizontal component Vx.
we know, Vx = vcostheta.
So, E = 1/2m(vcostheta)^2 = 1/2 m× v^2/2 = E/2